Puzzles with numbers and things

[Monique]

Round 276 - More Pinks and Browns

This one is harder. Charlie again goes and buys some balls (each pink or brown, equally likely, as before) from Barry; and Gordon then does too - but this time Gordon buys the same number of balls as Charlie.

In an act of kindness, Charlie then gives one of his balls, chosen at random, to Gordon.

Again, how likely is it that Gordon has more pinks to pot than Charlie, if...

(a) Charlie bought 3 balls from Barry, or

(b) (if you can) Charlie bought 147 balls (you can give this as an expression rather than calculating it).

Answers / guesses to either part on the thread please...

So at the end of the day Gordon has 2 ball more than Barry, randomly picked ...

the general formula giving the probability of Gordon getting more pinks than Charlie would be, if they both buy n balls to start with ...

P = (1/2^(2*n))*(Sum(i=0 to n-1; (n-1)!/(i!*(n-1-i)!*Sum(j=i+1 to n+1; (n+1)!/j!*(n+1-j)!))

for a) I get 42/64 ...

This number should decrease as n increases (converging to 1/2 by excess?). The fact that it's higher than 1/2 might look puzzling as the answer would be the same if the question was about browns, not pinks. But then Gordon can now have both more pinks and more browns than Charlie...

well ... I hope this is OK (you tell me about a brain teaser early morning after a snookery evening! )

[davis_greatest]

So at the end of the day Gordon has 2 ball more than Barry, randomly picked ...

the general formula giving the probability of Gordon getting more pinks than Charlie would be, if they both buy n balls to start with ...

P = (1/2^(2*n))*(Sum(i=0 to n-1; (n-1)!/(i!*(n-1-i)!*Sum(j=i+1 to n+1; (n+1)!/j!*(n+1-j)!))

for a) I get 42/64 ...

This number should decrease as n increases (converging to 1/2 by excess?). The fact that it's higher than 1/2 might look puzzling as the answer would be the same if the question was about browns, not pinks. But then Gordon can now have both more pinks and more browns than Charlie...

well ... I hope this is OK (you tell me about a brain teaser early morning after a snookery evening! )

Congratulations, Monique! This gives the right answer, although it can be simplified in a way to let you calculate it for bigger n.

Suppose they buy n balls each. So there are 2n balls. Charlie gives Gordon a ball, so Charlie then has n-1 and Gordon n+1.

There are 3 mutually exclusive possibilities:

(a) Gordon has more pinks than Charlie, and more browns. This means he must have exactly one more of each. In this case, number of Charlie pinks + number of Gordon browns must equal n; similarly, number of Charlie browns + number of Gordon pinks must equal n. (For example, if Charlie has k pinks and n-1-k browns, then Gordon would need to have n-k browns and k+1 pinks.)

(b) Gordon has more pinks than Charlie, but not more browns.

(c) Gordon has more browns than Charlie, but not more pinks.

(b) and (c) each have the same likelihood.
If the probability of possibility (a) is x, then the probability of (b) and of (c) is therefore each (1-x)/2.

We need to add possibilities (a) and (b), giving x + (1-x)/2 = (1+x)/2

For (a), we need the number of ways of choosing n balls from 2n: this is (2n)!/n!².
Each way of choosing 2n balls has probability (1/2)^2n.

So x= (2n)!/(n!².2^2n).

So the formula for the required probability is (1+x)/2 = [1+(2n)!/(n!².2^2n)]/2.

For n=3, this is [1+6!/(3!².2^6)]/2. = [1 + 20/64]/2 = 21/32, as you said!

For n= 147, it comes out at around 0.523.

Incidentally, as n becomes large, the probability becomes close to (1+ 1/√(n π))/2

where π = pi = 3.1415....

[davis_greatest]

Round 270 - Pooker

I have just thrashed Charlie during a frame of pooker - it's a sort of cross between snooker and pool. There are many balls, numbered from 1 upwards. So the 1-ball is worth 1 point, the 2-ball is worth 2 points etc. Balls must be potted in ascending order.

....

Round 277 - Another Pooker / Pool mix-up

Charlie is making another break at Pooker (see above), on a big table - let's say there are as many as 1 million balls. Referee Van Parkedoutside Jerhouse calls out the scores. If Charlie's break consisted, for example, of potting the 4-ball, 5-ball and 6-ball, then Van would call a break of 15 points (4+5+6).

On the adjacent table, I am playing 9-ball pool. However, referee Talia Mabb has got a bit mixed up about which game she is refereeing and she calls out breaks. However, instead of adding the points for each ball, she multiplies. So, if I potted 1-ball, 2-ball, 3-ball for example, she would call "six" (1 x 2 x 3), instead of "six" (1+2+3).

I break off at 9-ball and proceed to clear the whole table!

Amazingly, the final break that Talia calls is exactly the same as the break that Van is calling out on the adjacent table.

In how many different ways could Charlie have made his Pooker break?

Answers by... however you like.

[davis_greatest]

Just to clarify some points relating to round 277, in relation to some questions etc received:

In 9-ball pool, the balls are numbered 1, 2, 3, 4, 5, 6, 7, 8, 9.

In Pooker, balls must be potted consecutively in ascending order. So, if a break begins with the 4-ball, for instance, the next ball to be potted would be the 5-ball.

[davis_greatest]

Some first hints to round 277, for anyone trying to get on track. The hints give progressively more away - select the text to read.

Hint 1: ( Try looking for the number of ways that you can make consecutive integers, not necessarily positive, add up to any integer n. Then you will put n=9! )

Hint 2: ( If you have an odd number of consecutive integers (not necessarily positive) adding to n, they will be of the form:
(x-y) + .... + x + .... + (x+y) = ??? )

Hint 3: ( (x-y) + .... + x + .... + (x+y) = x(2y+1)
So the number of such sequences is equal to the number of positive *** divisors of n. )

Hint 4: ( The number of such sequences is equal to the number of positive odd divisors of n. )

Hint 5: ( ... and if you have an even number of consecutive integers (not necessarily positive) adding to n, they will be of the form:
(x-y+1) + .... + x + (x+1) + .... + (x+y) = ??? )

Hint 6: (
(x-y+1) + .... + x + (x+1) + .... + (x+y) = y(2x+1)
So the number of such sequences is equal to the number of positive *** divisors of n. )

Hint 7: ( The number of such sequences is equal to the number of positive odd divisors of n. )

Hint 8: ( So the total number of such sequences, with either an odd or even number of consecutive integers, is equal to twice the number of positive odd divisors of n.

Now you need to get rid of the ones that do not involve only positive integers. How? )

Hint 9: ( For each such sequence that includes some non-positive integers, there is exactly one sequence with only positive integers - and vice versa. Why?)

Hint 10: ( Therefore, the number of possible breaks is equal to the number of positive odd divisors of 9! = 1x2x3x4x5x6x7x8x9.

You're now halfway to solving the problem!

So how many positive odd divisors are there?)

Hint 11: ( You need to find the prime factorisation of 9!
)

[abextra]

I've got 20 positive odd divisors so far... ... actually I don't understand anything here anymore...

[davis_greatest]

I've got 20 positive odd divisors so far... ... actually I don't understand anything here anymore...

Well... congratulations abextra! 20 is the correct answer!

And congratulations also for turning green with your 300th post!

PS can you explain why there are 20 odd divisors?

[abextra]

Well... 20 is the correct answer!

PS can you explain why there are 20 odd divisors?

I have no idea WHY... I just multiplied the odd prime factors as many ways as I could find...

Your break was 9! = 362 880, odd prime factors of this number are 3, 3, 3, 3, 5, 7. I found 19 different combinations to multiply them and of course added number 1. So the odd divisors are

1, 3, 5, 7, 9, 15, 21, 27, 35, 45, 63, 81, 105, 135, 189, 315, 405, 567, 945 and 2835.

BTW, I think there are only 18 different ways to make a Pooker break of 362 880, the longest possible consists of 567 balls (357-ball - 923-ball).

[snookersfun]

...but then we still need to add a couple of even divisors

[davis_greatest]

I have no idea WHY... I just multiplied the odd prime factors as many ways as I could find...

Your break was 9! = 362 880, odd prime factors of this number are 3, 3, 3, 3, 5, 7. I found 19 different combinations to multiply them and of course added number 1. So the odd divisors are

1, 3, 5, 7, 9, 15, 21, 27, 35, 45, 63, 81, 105, 135, 189, 315, 405, 567, 945 and 2835.

BTW, I think there are only 18 different ways to make a Pooker break of 362 880, the longest possible consists of 567 balls (357-ball - 923-ball).

The prime factorisation of 9! is 9! = 2^7 x 3^4 x 5 x 7
so if we just look at the odd prime factors, they are 3^4 x 5^1 x 7^1.

We want the odd divisors of 9!, so we say:

the number of possible powers of 3 is 4+1 = 5 (i.e. powers 0, 1, 2, 3 or 4)
the number of possible powers of 5 is 1+1 = 2 (i.e. powers 0 or 1)
the number of possible powers of 7 is 1+1 = 2 (i.e. powers 0 or 1)

So the number of odd divisors of 9! is 5 x 2 x 2 = 20. 20 is therefore the answer to the question (the hints explain why).

There is no need to work out the Pooker breaks to answer the question. All you need to do is factorise 9! and then calculate 5 x 2 x 2 - but if you want the breaks, they are:

89 to 856
357 to 923
694 to 1098
995 to 1309
1290 to 1545
1826 to 2014
2621 to 2755
3404 to 3508
4440 to 4520
5729 to 5791
8042 to 8086
10351 to 10385
13427 to 13453
17270 to 17290
24185 to 24199
40316 to 40324
51837 to 51843
72574 to 72578
120959 to 120961
362880

These come out of Excel quite easily as each one corresponds to one of the 20 odd divisors of 9!.

[davis_greatest]

I have no idea WHY...

OK, another explanation of why each Pooker break of n corresponds to one of the positive odd divisors of n. (In this question, n was 1x2x3x4x5x6x7x8x9 but it could be anything.)

Suppose that s is an odd divisor of n. So we can write n = s.t (for some integer t>0).
As s is odd, we must be able to write s=2q+1 (q an integer >=0).

If t>q, then take the Pooker break beginning with ball t-q and ending with ball t+q. The average value of each ball is t and there are 2q+1 = s balls, so the break adds up to s.t = n.

If t<=q, then take the Pooker break beginning with ball q-t+1 and ending with ball t+q. The average value of each ball is (2q+1)/2 = s/2 and there 2t balls, so the break again adds up to (s/2).2t = s.t = n.

Hopefully that's a better explanation than my original hints!

[snookersfun]

The prime factorisation of 9! is 9! = 2^7 x 3^4 x 5 x 7
so if we just look at the odd prime factors, they are 3^4 x 5^1 x 7^1.

We want the odd divisors of 9!, so we say:

So the number of odd divisors of 9! is 5 x 2 x 2 = 20. 20 is therefore the answer to the question (the hints explain why).

There is no need to work out the Pooker breaks to answer the question. All you need to do is factorise 9! and then calculate 5 x 2 x 2 - but if you want the breaks, they are:


89 to 856 even divisor! 256x3
357 to 923
694 to 1098
995 to 1309
1290 to 1545 even divisor! 256= 2^8
1826 to 2014
2621 to 2755
3404 to 3508
4440 to 4520
5729 to 5791
8042 to 8086
10351 to 10385
13427 to 13453
17270 to 17290
24185 to 24199
40316 to 40324
51837 to 51843
72574 to 72578
120959 to 120961
362880

These come out of Excel quite easily as each one corresponds to one of the 20 odd divisors of 9!.

correct me if I am wrong. Haven't we kind of ignored the issue of the even divisors so far (well there is some in previous post)? I highlighted the 2 in the above list kind of fortunately in this case (I think, not claiming to have thought this through until the bitter end) replacing the two odd divisors (3x3x3x3x7x9 and 3x3x3x5x7), which would be too high to form positive results.

But assume now your number would be just a factor of 4 bigger (or actually go for any high enough multiple of 2), now, your odd factors are still the same, moreover they will now be all allowed (in the case of 4 one more will be), as the 'averages' are now all possible (i.e. all lead to positive results).
So, at one stage we will have all those 20 possible results + need to add to that 2 possible even divisors.
Those even divisors will be 2^(n+1) (with n= number of 2 divisors), as this will lead to an average of ?????.5 and in the case of the above number also 3 multiplied by that number.
So, yes, they do correspond as there are incidentally a possible 20 even factors as well (2^(n+1) times 1,3,5, etc...). One is still left with weeding out the positive/negative ones though, isn't one? (which the previous post just might explain somehow)

[davis_greatest]

correct me if I am wrong. Haven't we kind of ignored the issue of the even divisors so far (well there is some in previous post)? I highlighted the 2 in the above list kind

I'm not sure what you mean by ignored the issue of even divisors? You only need to count odd divisors, as each break corresponds to one such divisor. The two you have highlighted are indeed different from the other 18, since they correspond to a break where there is an even number of balls (the other 18 have an odd number of balls) - these two correspond to the case t<=q using the notation in post 2560.

For example,

- the break 89 to 856 corresponds to the odd divisor s=3³ x 5 x 7 = 945, and q=472. Here, t=384 (since 362880 / 945 = 384), so the break begins at q-t+1 = 89.

- the break 1290 to 1545 corresponds to the odd divisor s=3^4 x 5 x 7 = 2835, and q=1417. Here, t=128, so the break begins at q-t+1 = 1290.

But assume now your number would be just a factor of 4 bigger...

That wouldn't make any difference to the number of breaks, as there would would still be the same number (20) of odd factors. So, if n=s.t, every odd factor s would be the same as before and every t would be 4x bigger than before. There would still be 20 breaks, but they would be different from before.

[April madness]

ehm... wot does that all mean?

[PaulTheSoave]

SF is still full of adrenaline after last night! Just ignore this thread for a couple of hours.

ehm... wot does that all mean?